Problem: Solve for $t$. $4\left(t+\dfrac14\right) = 3$
Let's divide and then subtract to get $t$ by itself. $\begin{aligned}4\left(t+\dfrac14\right) &= 3 \\ \\\\ \dfrac{4\left(t+\dfrac14\right)}{{4}} &= \dfrac{3}{{4}} ~~~~~~~\text{divide each side by } {4}\\ \\ \dfrac{\cancel{4}\left(t+\dfrac14\right)}{\cancel{{4}}} &= \dfrac{3}{{4}} \\ \\ t +\dfrac14&= \dfrac{3}{{4}} \end{aligned}$ $\begin{aligned} t +\dfrac14&= \dfrac34 \\ \\ t+\dfrac14 {-\dfrac14}&= \dfrac34{-\dfrac14}~~~~{\text{subtract }\dfrac14} \text{ from each side} \text{ to get } t \text{ by itself }\\ \\ t+\cancel{ \dfrac14} {{-}\cancel{{\dfrac14}}}&= \dfrac34{-\dfrac14}\\ \\ t &=\dfrac34{-\dfrac14}\end{aligned}$ The answer: $t={\dfrac12}$ Let's check to make sure. $\begin{aligned} 4\left(t+\dfrac14\right)& = 3 \\\\ 4\left({\dfrac12}+\dfrac14\right) &\stackrel{?}{=} 3 \\\\ 4\left(\dfrac34\right)&\stackrel{?}{=} 3 \\\\ 3 &= 3 ~~~~~~~~~~\text{Yes!} \end{aligned}$